JEE Main 2026 · 6 Apr · Shift 2 · Mathematics Q18

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MathematicsSingle correct+4 / -1Question 18 of 75

Let $f:\mathbf R\to\mathbf R$ be such that $f(xy)=f(x)f(y)$ , for all $x,y\in\mathbf R$ and $f(0)\ne0$ . Let $g:[1,\infty)\to\mathbf R$ be a differentiable function such that
$x^2g(x)=\int_1^x\left(t^2f(t)-tg(t)\right)dt.$
Then $g(2)$ is equal to :

\dfrac{13}8

\dfrac{11}{16}

\dfrac{15}{32}

\dfrac{17}{64}

Answer and worked solution

Official answer: C

f. Putting $x=0$ : $f(0)=f(0)f(y)$ , and $f(0)\ne0$ , so $f\equiv1$ .

Differential equation. Differentiating,
$2xg+x^2g'=x^2-xg\ \Rightarrow\ g'+\frac3xg=1.$

Solve. $(x^3g)'=x^3$ , so $x^3g=\tfrac{x^4}4+C$ . At $x=1$ , $g(1)=0$ , so $C=-\tfrac14$ .

Value. $g(2)=\dfrac{4-\frac14}8=\dfrac{15}{32}$ .

(C)

Topics:Linear first-order equations Fundamental theorem of calculus

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JEE Main 2026 · 6 Apr · Shift 2, Mathematics question 18